A uniform current-carrying ring of mass m and | vedclass

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(A) For the ring to remain in a horizontal position,the net torque about the point of suspension must be zero.The gravitational force $mg$ acts at the

Vedclass · Accepted Answer

$\frac{mg}{\pi R B_0}$

Vedclass · Answer

(A) For the ring to remain in a horizontal position,the net torque about the point of suspension must be zero.The gravitational force $mg$ acts at the center of the ring,creating a torque $	au_g = mgR$ about the suspension point.The magnetic field $B_0$ exerts a magnetic torque on the current-carrying ring. The magnetic dipole moment of the ring is $M = I A = I(\pi R^2)$.The magnetic torque is given by $	au_m = M B_0 = I(\pi R^2) B_0$.Equating the torques for equilibrium: $	au_g = 	au_m$.$mgR = I(\pi R^2) B_0$.Solving for $I$: $I = \frac{mgR}{\pi R^2 B_0} = \frac{mg}{\pi R B_0}$.

$A$ uniform current-carrying ring of mass $m$ and radius $R$ is suspended by a massless string as shown. $A$ uniform magnetic field $B_0$ exists in the region to keep the ring in a horizontal position. Determine the current $I$ in the ring.

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